JC-NRLF 


273 


AND 
,TR  I  SECTION 

OF 

.ANGLES 


LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 





Dividing 
Angles 

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I                  1 

1 

|      J.  C,   WILLMON      | 
&                                          5 

£                                          a 
*                                                   3 
*  ittitoittifc*  xivi/  *****  ifcifcitoVfc  H 

LOS    ANGELES,         CALIFORNIA 
— 19O4— 

COPYRIGHTED  1904  3 


PREFACE 


To  bisect  an  angle  and  to  divide  it  into  multiples 
of  two  is  easy;  to  trisect  an  angle  is  much  more 
difficult,  and  a  rule  that  will  divide  angles  into  ANY 
number  of  equal  parts  has  long  been  sought  for,  but 
has  never,  so  far  as  I  know  been  heretofore  discovered. 
The  problem  was  supposed  to  be  impossible,  and  prob- 
able is  impossible  by  means  of  Elementary  Geometry. 
The  line  i  e  f  k  is  a  slightly  curved  line,  therefore 
the  solution  does  not  come  within  the  domain  of 
Euclidian  or  Elementary  Geometry.  In  "Higher  Ge- 
ometry" it  is  allowable  to  solve  problems  by  means 
of  curved  lines  drawn  through  a  series  of  fixed  points 
and  my  solution  of  dividing  angles  as  herein  given 
comes  within  the  domain  of  "Higher  Geometry",  and 
I  believe  the  solution  to  be  new  and  correct. 

J.  C.  WILLMON, 

Los  Angeles,  Cal. 


August  9,  1904. 


PROBLEM 


To  divide  any  angle  into  any  number  of  equal  angles: 
Let  b  a  c  be  the  given  angle  which  it  is  proposed 
to   divide    into    any  number    of   equal  parts,  as  for 
example,  five. 

Take  a  point  1  on  the  line  a  c  make  a  1  =  1,  2  = 
2,  3  =  3,  4  =  4,  5.  With  a  as  center  and  a  1  as 
radius  describe  the  arc  1  d.  With  a  as  center  and  a 
5  as  radius  describe  the  arc  5  k  j.  With  a  as  center 
describe  a  number  of  indefinite  arcs,  as  e  e,  i  i  &c. 
With  a  radius  =  1  d  mark  5  points  on  the  arcs  5  k 
j,  e  e,  i  i,  &c;  then. 

5  k  =  5.  1  d 
e  e  =  5,  1  d 
i  i  =  5,  1  d 

Through  the  points  i  e  k  draw  the  slightly  curved 
line  i  e  f  k  cutting  the  line  b  a  at  f . 

With  a  as  center  and  a  f  as  radius  describe  the 
arc  f  g  h.  Make  h  g  =  I  d.  Then  f  h  =  5  1  d  and 
h  g  =  1  d  and  the  angle  g  a  h  =  ]-5  angle  b  a  c. 

PROOF 

5  k  =  5    1  d  by  construction, 
e  e  =  5    i  d  by  construction, 
i  i  =  5   1  d  by  construction, 
fh=5k=ee=ii  by  construction. 
.".  f  h  =  5  1  d  by  construction, 
h  g  =  1  d  by  construction. 
fh=51d=5hgby  construction, 
h  g  =  1-5  f  h  by  construction. 
/.  angle  g  a  c  =        1-5  angle  b  a  c. 


THE 

SECRET  OF  THE  CIRCLE 


AND 


TRISECTION  OF  ANGLES 


BY 


JEREMY  CARLISLE  WILLMON 


McBRIDE    PRESS 
316  W.  SECOND  STREET,    Los  ANGELES,   CAL. 


4 


COPYRIGHT   BY 

JEREMY  C. 

1903 


R  A 

OF  TH£ 


UNIVERSITY 


PREFACE. 


The  most  famous  geometrical  and  mathematical 
problem  of  the  ages  is  the  squaring  of  the  circle. 

The  problems  that  have  heretofore  defied  solu- 
tion by  fixed  rule  or  method  are : 

i st.  The  construction  of  a  square  that  shall 
exactly  equal  in  area  any  given  circle. 

2nd.  The  construction  of  a  circle  that  shall 
exactly  equal  in  area  any  given  square. 

3rd.  The  construction  of  a  straight  line  that 
shall  exactly  equal  the  circumference  of  any  given 
circle. 

4th.  The  construction  of  a  circle  the  circumfer- 
ence of  which  shall  exactly  equal  any  given  straight 
line. 

The  circumference  of  a  circle  has  been  proven  by 
mathematical  calculations,  and  the  ratio  of  its  circum- 
ference to  its  diameter  has  been  computed  to  400 
decimal  places. 

5 

124491 


To  reduce  a  circle  to  a  square  or  a  square  to  a 
circle,  or  to  reduce  the  circumference  of  a  circle  to  a 
straight  line,  or  a  straight  line  to  the  circumference  of 
a  circle,  has  always,  in  the  past,  required  complicated 
mathematical  calculations,  always  with  chances  of 
making  mistakes  in  figures,  and  the  results  at  best 
only  *  'approximate."  There  may  have  been  many 
"approximate"  solutions  of  the  above  problems. 

Mathematical  calculations  serve  to  prove  the 
length  of  a  straight  line,  when  properly  constructed, 
that  shall  equal  the  circumference  of  a  circle,  but  the 
calculations  do  not  and  cannot  make  the  line. 

Let  it  be  conceded  that  the  circumference  of  a 
circle  may  be  measured  by  making  a  wheel  or  disc  of 
the  same  diameter  and  circumference  as  the  circle, 
but  to  do  this  for  every  circle  to  be  calculated  would 
be  utterly  impracticable. 

These  four  problems  are  all  solved  by  the  right- 
angled  triangle  ABC,  Figure  i. 

By  mathematical  calculations  the  results  are 
proven  "correct  to  infinity." 

The  triangle  is  to  be  constructed  according  to  the 
geometrical  diagram  a  b  c,  Figure  2. 

It  is  to  be  made  of  metal  or  other  material^  and 


may  be  of  any  size.  For  the  sake  of  simplicity  I  have 
adopted  the  following:  AtC  =  10  inches;  A  B  = 
8.862269254-!-  inches  ;  B  C  =  4.6325138  +  inches. 
Then  BAG  forms  an  angle  of  27°  36' — .  Whatever 
the  size  of  the  triangle  may  be,  the  proportions  between 
the  parts  are  always  the  same.  In  geometrical  prob- 
lems the  proofs  must  be,  not  "approximate,"  but  ex- 
act. The  inch,  the  foot,  the  metre  and  other  stand- 
ards of  measurement  may  not  be  exactly  the  same  ten 
thousand  years  from  now  as  they  are  today  ;  but  math- 
ematical and  geometrical  truths  are  the  same  always, 
unchangeable  and  eternal.  Therefore  the  triangle  ABC, 
Figures  i  and  2,  as  a  standard  of  measurement  between 
circles,  squares,  straight  lines  and  circumferences  may 
be  made  of  any  size,  anywhere  at  any  time,  and  the 
angle  being  always  the  same  and  the  triangle  being 
always  similar,  is  a  geometrical  truth,  "the  same  yes- 
terday and  today  and  forever." 

Let  it  be  conceded  : 

ist.  That  it  is  as  easy  to  construct  a  circle  as  it 
is  to  construct  a  straight  line. 

2nd.  That  it  is  as  easy  to  construct  a  mechanical 
wheel  or  disc  as  it  is  to  construct  a  mechanical  square. 

3rd.     That  a  wheel  or  disc  may  be  revolved  along 


a  plane  surface  in  a  straight  line,  and  that  a  straight 
line  may  be  constructed  ^and  marked  equal  to  the 
perimeter  of  the  wheel  or  disc  and  equal  to  the  cir- 
cumference of  a  circle  of  the  same  diameter. 

4th.  That  this  line  may  be  divided  into  4  equal 
parts. 

5th.  That  the  geometrical  diagram  a  b  c,  Figure 
2,  may  be  constructed  and  used  as  a  standard  for  con- 
structing similar  angles  and  triangles  and  the  right- 
angled  Triangle  A  B  C,  Figure  i,  may  be  constructed 
with  the  angle  BAG  similar  to  angle  b  a  c  in  Figure 
2.  Then  with  the  triangle  a  square  may  be  con- 
structed equal  in  area  to  any  given  circle  ;  a  circle 
may  be  constructed  equal  in  area  to  any  given  square  ; 
a  straight  line  may  be  constructed  equal  to  the  cir- 
cumference of  any  given  circle  ;  and  a  circle  may  be 
constructed  whose  circumference  equals  any  given 
straight  line. 

Therefore  the  4  problems  are  solved. 

Another  celebrated  problem  is  the  trisection  of 
angles  by  elementary  geometry  (straight  lines  and  cir- 
cles). It  was  attempted  in  the  schools  of  the  ancients, 
and  has  probably  been  attempted  by  most  students 
and  professors  of  geometry.  It  has,  I  am  informed, 

8 


been  solved  by  "Higher  Geometry"  by  means  of  an 
"auxiliary  curve,"  but  I  have  not  seen  the  solution. 
I  include  the  problem  herein,  a  simple  method  by  ele- 
mentary geometry,  trusting  it  may  prove  interesting 
to  the  reader. 

To  anyone  finding  objections  to  my  methods  of 
solving  any  of  the  foregoing  problems,  I  would  say 
that  I  am  willing  to  have  my  solutions  of  any  or  all  of 
them  relegated  to  obscurity  when  better  methods  are 
discovered  for  accomplishing  the  same  purposes. 

I  have  read  that  these  problems  are  impossible, 
and  that  none  but  ignorant  persons  would  waste  time 
attempting  to  solve  them.  I  will  confess  that  my 
schooling  did  not  include  geometry  nor  any  of  the 
higher  branches  of  learning  ;  and  therefore  it  may  be 
well  for  me  to  explain  how  I  came  to  solve  the  prob- 
lems. 

It  all  came  about  from  making  an  analysis  of  the 
poem  "In  the  Distance."  Noticing  that  it  contained 
7  verses  and  7  lines  to  the  verse  led  me  to  looking  for 
further  coincidences,  and  through  this  my  attention 
was  attracted  to  triangular  numbers  and  triangles. 
The  idea  came  to  me  that  through  this  poem  would  be 
solved  the  secret  of  the  circle,  and  I  turned  my  atten- 

9 


tion  to  the  same,  with  the  results  as  herein  given. 
The  poem  having  led  to  the  solution  of  the  problems 
under  consideration,  it  is  published  in  connection  here- 
with, together  with  a  number  of  the  coincidences  ; 
they  may  lead  to  the  solution  of  other  problems. 
None  of  the  coincidences  were  noticed  or  discovered 
until  the  poem 'was  complete,  word  for  word  and  letter 
for  letter  as  it  now  is  ;  not  a  word  was  changed  or 
abbreviated  to  bring  about  any  of  these  results.  I  do 
not  attempt  to  explain  the  coincidences.  I  have 
endeavored  to  demonstrate  the  facts  and  truths  herein, 
only  as  I  found  them. 

J.  C.  WILLMON. 
Los  ANGELES,  CAL.,  July  13,  1903. 


10 


Figure     1 


Figure    2 


11 


Figure  4 


13 


C  1 


Figure  6 


15 


Figure    T 


I. 

Triangle  ABC,  Figure  i ,  is  a  mechanical  triangle 
constructed  similar  to  geometrical  triangle  a  b  c, 
Figure  2. 

II. 

Construct  a  right  triangle  A  B  C  so  that  A  C  = 
10 ;  A  B  =  8.862269254  +  ;  B  C  —  4.6325138  +,  or 
in  the  same  proportions. 

Figure  2.  Take  any  circle  a  k  c  f ;  construct  a 
round  wheel  or  disc  of  the  same  diameter  as  the  circle 
a  k  c  f ;  roll  the  wheel  or  disc  one  revolution  on  a  plane 
surface  in  a  straight  line  f  j  ;  then  the  line  f  j  =  the 
circumference  of  the  circle  a  k  c  f .  Divide  f  j  into  4 
equal  parts  ;  let  f  g  =  ^  f  j.  Draw  the  diameters  k  f 
and  a  c  I  to  each  other  ;  make  a  d  =  f  g  ;  erect  the 
]_db  cutting  the  circumference  at  b ;  connect  b  and  a 
and  b  and  c.  Then  triangle  a  b  c  is  the  triangle  re- 
quired. 

THEOREM. 

Figure  2.  In  any  circle  a  k  c  f ,  if  a  d  on  the 
diameter  a  c  be  made  =  %  of  the  circumference  of  the 
circle  and  the  line  d  b  be  erected  _L  to  a  c,  cutting  the 
circumference  of  the  circle  at  b,  then  a  b  —  one  side  of 
a  square  containing  the  same  area  as  the  circle.  Con- 
versely, if  a  b  be  made  =  to  one  side  of  a  square  con- 

19 


taining  the  same  area  as  the  circle,  and  the  line  b  d  be 
made  _J_.to  the  diameter  a  c,  cutting  the  diameter  at  d, 
then  a  d  =  %  circumference  of  the  circle  a  k  c  f . 

PROOF. 

a  b  c  is  a  right  A  being  inscribed  in  a  semi-circle. 

Suppose  the"  radius  a  e  =  5,  then  a  c  =  10. 
Then  f  j  =  31.415926535897932  +  =  circumference  by 
by  construction  ;  f  g  =  ^  :  f  j  —  7.853981633974483  +  - 

a  d  =  f  g  by  construction. 

e  d  =  a  d  —  ae  =  2.853981633974483  +. 

dc  =  ec  —  e  d  =  2.146018366025516  +  . 

e  b  =  radius  =  5  by  construction. 

b  d  =  i/  e  b2  —  e  d2  =  4.1054584  +  Ref  I. 

a  b  —  j/  a  d2  +  b  d2  =  8.862269254+  Ref.  II. 

b  c  =  ]/dc2  +  bd2  =4.6325138  +  Ref.  III. 

.*.  a  c  =  10  ;  a  b  =  8.862269254  +  ;  be  ==  4.63- 
25138+  Q.  E.  D. 

Ref.  I  b  d  is  a  leg  of  a  right  A  e  b  d. 

Ref.  II  a  b  is  the  hypothenuse  of  right  A  a  b  d. 

Ref.  Ill  b  c  is  the  hypothenuse  of  right  Abed. 

PROBLEM  I. 

To  construct  a  square  equal  in  area  to  any  given 
circle. 

Construction  :  Take  any  circle  a  b  c  g,  Figure 
3  ;  draw  diameter  a  e  c  ;  construct  on  the  diameter 

20 


a  ec  the  A  a  b  c  similar  to  A  A  B  C,  Figure  i.  On 
a  b  construct  the  square  a  b  d  f ,  which  is  the  required 
square. 

PROOF. 

Suppose  the  diameter  a  e  c  =  6  ;  A  abcis  simi- 
lar to  A  A  B  C,  Figure  i  by  construction. 

.'.  ab  =  ^  of  AB,  Figure  i  =  5.3173615524+. 

5.3173615524  +'2  28.2743338823+  :  area 

square  a  b  d  f . 

Diameter  a  e  c  =  6.  .'.  area  of  circle  a  b  c  g  — 
28.2743338823  +. 

.'.  The  square  constructed  on  a  b  =  circle  = 
a  b  c  g.  Q.  K.  F.  • 

PROBLEM  II. 

To  construct  a  circle  that  shall  equal  in  area  any 
given  square. 

Construction  :  Take  any  square  a  b  d  g,  Figure 
4  ;  on  a  b  construct  A  a  b  c  and  a  b  f ,  similar  to  A 
ABC,  Figure  i  ;  the  lines  a  c  and  b  f,  cutting  or 
being  produced  until  they  cut  the  sides  of  the  square 
at  c  and  f,  with  e  as  center  and  e  a  as  radius,  describe 
the  circle  a  b  c  f ,  which  is  the  required  circle. 

PROOF. 

Suppose  the  side  a  b  =  6  and  the  area  =  36. 

a  c  and  b  f  are  equal ;  they  bisect  each  other  at  e. 

2L 


.'.  ae  —  eb  =  ef  —  ec=  radius  of  circle  a  b  e  f. 

A  a  be,  Figure  4,  is  similar  to  A  A  B  C,  Figure  i , 
by  construction  .'.  the  sides  are  proportional. 

.'.   If  a  b  =  6,  a  c  =  6.770275  +. 

a  e  =  y2  a  c  —  3.3851375  +  —  radius  of  circle 
a  b  c  f . 

Circumference  X  \  radius  =  area  of  circle. 

.*.  area  of  circle  —  36. 

.*.  circle  a  b  c  f  =  square  a  b  d  g.          Q.  E.  F. 

PROBLEM  III. 

To  construct  a  line  equal  to  the  circumference  of 
a  given  circle. 

Construction  :  Take  any  circle  a  b  c  h,  Figure  5  ; 
draw  diameter  a  c  ;  on  a  c  construct  A  a  b  c,  similar 
to  A  A  B  C,  Figure  i.  From  b  drop  the  _j_  b  d  ;  pro- 
duce a  c  to  g,  making  a  g  =  4  a  d  ;  then  a  g  =  the 
circumference  of  the  circle  a  b  c  h. 

PROOF. 

Suppose  a  c  =  6. 

In  Figure  2,  a  d  =  7.8539816  +. 

In  Figure  5,  triangle  a  b  d  is  similar  to  triangle 
a  bd,  Figure  2. 

. '.  Therefore  a  d  Figure  5  =  T%  a  d  Figure  2  = 
4.71238898  +  . 

4.71238898  +  X  4  =  18.84955592  +  :=  a  g. 

a  c  =  6  .'.  circumference  a  b  c  h=  18.849- 
55592+. 

.-.  a  g  —  circumference.  Q.  E.  F. 


PROBLEM  IV. 

To  construct  a  circle,  the  circumference  of  which 
shall  equal  any  given  straight  line. 

Construction :  Take  any  straight  line  a  g,  Figure 
6  ;  divide  a  g  into  4  equal  parts  ;  let  a  d  =  ^  a  g; 
at  d  erect  the  _[_  d  b  ;  at  a  construct  angle  c  a  b  — 
angle  CAB,  Figure  i ;  the  side  of  angle  cab  cutting 
or  being  produced  until  it  cuts  d  b  at  b  ;  draw  j  e  the 
perpendicular  bisector  of  a  b  cutting  a  g  at  e  ;  with  e 
as  center  and  e  a  and  e  b  as  radius  describe  the  circle 
a  b  c  f ,  which  is  the  circle  required. 

The  line,  circle  and  angle  of  problem  4  being 
similar  to  the  lines,,  circles  and  angles  of  problems  I, 
II  and  III,  the  reader  may  apply  the  proof. 

PROBLEM  V. 
To   Trisect  an  Angle. 

Take  any  angle  bac,  Figure  7,  not  greater  than  a 
right  angle  ;  take  any  point,  e  on  the  line  a  b  and 
draw  the  line  e  d  ||  a  c ;  with  e  as  center  and  e  a  as 
radius  describe  the  arc  a  g  f  h ;  draw  the  line  a  d  so 
that  f  d  =  e  a  ;  draw  f  i  ||  d  e  ;  with  a  as  center  and 
a  d  as  radius  describe  the  arc  c  d  j  b  ;  then  angle  d  a  c 

=  Y$  angle  bac. 

PROOF. 

d  f  =  e  f  by  construction  .'.  angle  e  d  f  =  angle 
d  e  f. 


Angle  c  a  d  =  angle  a  d  e.     Ref.  I. 

Angle  a  d  e  =  angle  d  e  f .     Ref.  II. 

Angle  e  d  f  =  angle  e  f  i.     Ref.  I. 

Angle  a  d  e  +  angle  d  e  f  +  angle  d  f  e  =  a 
straight  angle.  Ref.  III. 

But  angle  a  f  i  +  angle  e  f  i  H-  angle  d  f  e  =  a 
straight  angle. 

.'.  Angle  a  f  i  -f-  angle  e  f  i  =  angle  a  d  e  + 
angle  d  e  f . 

But  angle  d  e  f  =  angle  e  f  i  =  angle  a  d  e. 

.*.  a  f  i  =  a  d  e  =  c  a  d. 

.'.  Angle  a  f  e  =  2  angle  cad. 

But  angle  a  f  e  =  angle  d  a  b.     Ref.  II. 

.*.  Angle  d  a  b  =  2  angle  cad. 

.'.  Angle  c  a  d  =  ft  angle  b  a  c. 

Q.  E.  F. 

Ref.  I.  When  two  ||  lines  are  cut  by  a  third  line 
the  opposite  interior  angles  are  equal. 

Ref.  II.  In  an  isosceles  A  the  angles  opposite 
the  equal  sides  are  equal. 

Ref.  III.  The  sum  of  the  angles  of  a  A  is  equal 
to  a  straight  angle. 


NOTE. — The  triangle,  Figure  i,  may  be  found 
useful  in  solving  many  additional  problems,  notably 
the  surfaces  and  contents  of  spheres,  cylinders  and 
cones,  as  well  as  the  converse  of  the  problems. 

24 


IN  THE  DISTANCE. 


The  countless  leg-ions  passed  away, 
And  all  the  hosts  on  earth  today, 
L,ike  vanished  dreams  may  be  forgot, 
Their  names  and  deeds  remembered  not, 

Their  gilded  glories  gone  ; 
Their  works  as  rust  and  desert  dust, 

Fame's  phantom  shadows  flown. 

II 

Or  like  enchanted  music  rung1, 
Our  songs  attuned  to  cadence  sung1, 
Or  names  by  nrystic  fate  renowned, 
By  glamoured  ancient  glories  crowned, 

With  all  that  fame  endears, 
It  naught  would  be  to  you  or  me, 

Far  down  the  distant  years. 

Ill 

A  few  at  most  our  troublous  days  ; 
Unto  the  vast  unknown  we  gaze  ; 
A  glimmer  of  immortal  dawn, 
A  star  of  hope  still  shining  on, 

Gleams  through  the  darkest  sky  ; 
A  trust  that  good  shall  cross  the  flood, 

And  only  evil  die. 
25 


IV 

Where  doubt  exists  a  hope  may  live  ; 
None  know  the  gif ts  that  time  may  give  ; 
Above  our  highest  hopes  and  far 
Beyond  the  dreamer's  brightest  star, 

Have  faith!  for  us  may  rise 
The  future's  dawn,  the  shores  unknown, 

The  fadeless  lOden  skies. 

V 

Ivet  patience  ever  shield  thy  breast 
From  storm-tossed  waves  of  wild  unrest, 
And  love  make  all  thy  pathways  bright, 
Contentment  make  thy  burdens  light ; 

Ivet  gloomy  thoughts  forlorn, 
And  griefs  and  fears,  the  pains  and  tears, 

All  pass  like  mists  of  morn. 

VI 

Haste  not  to  leap  the  fabled  stream ; 
What  waits  beyond  we  may  not  dream  ; 
Rejoice  today,  yet  meekly  trust, 
That  only  good  above  our  dust, 

By  fate,  somewhere,  somehow, 
From  acts  of  ours  may  grow  as  flowers, 

In  far-off  years  from  now. 

VII 

Trust  not  in  fame  nor  wealth  to  bless  ; 
Go  help  the  poor  and  soothe  distress ; 
Be  brave,  be  true  and  do  your  best ; 
Do  good  until  with  God  you  rest, 

In  some  far  wondrous  home, 
And  all  will  be  as  well  with  thee, 
Through  all  the  years  to  come. 
26 


COINCIDENCES. 

There  are  three  words  in  the  title  and  300  words 
in  the  7  verses  ;  13  letters  in  the  title  and  1300  letters 
in  the  complete  work  ;  there  are  1287  letters  in  the  7 
verses,  not  counting  the  13  letters  in  the  title. 

There  are  7  verses,  7  lines  to  the  verse,  7  sentences 
and  7  periods.  There  are  7  times  7  words  in  the  7th  . 
verse.  There  are  also  coincidences  corresponding  to 
the  number  of  minutes  to  the  hour,  the  number  of 
hours  to  the  day,  and  the  number  of  months  in  the 
year.  There  are  as  many  syllables  to  the  verse  (52) 
as  there  are  weeks  in  the  year,  and  52  punctuation 
marks  are  used  in  the  7  verses. 

There  are  365  syllables  in  the  7  verses.  7X52  — 
364.  (The  extra  syllable  is  in  the  line  erding  with 
the  word  flowers.)  Also,  the  second  and  fourth  verses 
combined  have  365  letters,  and  the  fourth  and  sixth 
verses  combined  have  365  letters,  corresponding  to  the 
number  of  days  in  one  year. 

Only  one  exclamation  point  is  used  ;  it  divides  the 
words  exactly  ;  150  words  preceding  it  and  150  follow- 
ing it. 

The  line  numbers  of  the  third  verse,  15  +  16  +  17 
+  18  +  19+20+21  =  126-^-3  the  number  of  the  verse, 
equals  42,  the  number  of  words  in  the  verse.  The 
seventh  line  of  the  third  verse  is  the  shortest  line  in 
the  work  ;  there  are  three  times  as  many  words  in  the 

27 


verse  as  there  are  letters  in  this  line.  There  are  13  X 
1 3—  1 69  letters  in  the  verse;  the  sum  of  all  numbers 
from  i  to  169  equals  14365.  The  two  left-hand  figures 
will  represent  the  number  of  letters  in  the  shortest 
line,  and  the  three  right-hand  figures  the  number  of 
syllables  in  the  verses  and  the  number  of  days  in 
the  year. 

The  first  letter  of  the  alphabet  is  used  as  a  word 
and  for  the  commencement  of  words  33  times  ;  33 
commas  are  used  ;  there  are  33  letters  in  the  longest 
line  and  33  lines  preceding  it. 

There  are  29  letters  in  the  first  line,  29  letters  in 
the  last  word  of  each  line  of  the  first  verse  combined, 
29  letters  in  the  last  word  of  the  middle  line  of  each 
verse  combined  and  29  letters  in  the  last  word  of  each 
verse  combined. 

There  are  24  letters  in  the  last  line  and  24  letters 
in  the  first  word  of  each  verse  combined ;  the  sum  of 
all  numbers  from  i  to  24  =  300,  the  number  of  words 
in  the  7  verses. 

The  number  of  letters  in  the  alphabet  26,  multi- 
plied by  the  number  of  verses  7  =  182,  the  number  of 
letters  in  the  7th  verse.  Seven  lines  have  each  26  let- 
ters, viz:  2d,  i5th,  i9th,  24th,  36th,  38th  and  45th; 
these  numbers  added  together,  2  +  15  +  19+24+36+ 
38+45  =  179,  the  number  of  letters  in  the  (2-6-)  2nd 
and  6th  verses.  The  sum  of  all  numbers  from  i  to  26 
=  351,  the  number  of  letters  in  the  3d  and  7th  verses 
combined. 

28 


The  most  wonderful  of  all  numbers  is  1287.  The 
number  of  verses  7,  multiplied  by  the  number  of  let- 
ters 1287  =  9009;  the  answer  reads  the  same  either 
way  backward  or  forward.  1287  with  the  right-hand 
figure  transposed  and  added  to  the  left  results  in  828, 
which  reads  the  same  either  way  backward  or  forward. 

The  sum  of  all  numbers  from  i  to  1287  =  828828, 
which  reads  the  same  either  way  backward  or  forward, 
divided  in  the  middle  either  half  reads  the  same  either 
way  ;  828828  divided  by  2  and  the  answer  divided  in 
the  middle,  either  half  is  the  same  and  reads  the  same 
either  way.  In  all  the  series  of  triangular  numbers, 
the  number  1287  has  no  parallel;  no  other  number 
will  produce  parallel  or  similar  results.  In  the  last 
half  (150)  of  the  words  are  625  letters;  1287  -f  625 
—  1912.  The  sum  of  all  numbers  from  i  to  1912  = 
1828828. 

(To  find  the  sum  of  all  numbers  from  i  up  to  any 
number  required,  square  the  greater  number;  to  the 
product  add  the  greater  number  and  divide  by  2. 
Example:  10  X  10  =  100  +  10  =  no  ~-  2  =•  55, 
the  sum  of  all  numbers  from  i  to  10  ) 


1287  X  in  = 

1287  X  222  = 

1287  X  333  = 
1287  X  444  = 
1287  X  555  = 
1287  X  666  = 

29 

142857 
285714 
428571 
57H28 
7H285 
857H2 

Each  answer  contains  the  same  figures  in  the 
same  order,  but  beginning  at  a  different  place.  Multi- 
plied by  259  the  answer  is  six  threes  (333333).  Mul- 
tiplied by  518  the  answer  is  six  sixes  (666666).  Mul- 
tiplied by  a  trinity  of  sevens  (777)  the  answer  is  six 
nines  (999999)- 

The  decimal  of  \  is  .142857.  1000000  -r-  7  — 
142857.142857. 

The  decimal  of  Tyy  is  .001287.  IOOQOOO  -*-  777 
=  1287.001287. 

The  decimal  of  1287  is  .000777.  1000000  -8-  1287 
=  777.000777. 


MYSTIC  NUMBERS. 

(5)  (3)  5th  verse  3d  line  :  5X3  =  T5-  The  sum 
of  all  numbers  from  i  to  5  equals  15.  "Love"  is  the 
1 5th  word  of  the  5th  verse  in  the  3d  line,  and  the 
1 79th  word  of  the  work.  (7)  (4)  7th  verse  4th  line  : 
7  X  4  =  28.  The  sum  of  all  numbers  from  i  to  7  = 
28.  "God"  is  the  28th  word  of  the  7th  verse  in  the 
4th  line,  and  the  279th  word  of  the  work. 

(10).  The  sum  of  all  numbers  from  i  to  10  = 
55  ;  "Mystic"  is  the  55th  word  of  the  work  in  the 
loth  line. 

Commencing  with  the  Sun  as  i,  Mercury  2, 
Venus  3,  Earth  4,  Mars*5,  The  Asteroids  6,  Jupiter  7, 
Saturn  8,  Uranus  9,  Neptune  io,_Comets  n,  the  Fixed 


Stars  and  Nebula  12,  and  i3th  the  Unknown  ;  13 
multiplied  by  the  number  of  verses,  13  X  7  =  91. 
''Unknown"  is  the  gist  word  of  the  work. 

The  word  numbers  added,  55+91  +  179  +  279 
=  604.  The  figures  of  the  answer  added  separately, 
6+0  +  4=1  o.  The  sum  of  all  numbers  from  i  to 
10  =  55.  The  figures  in  the  numbers  of  the  words 
added  separately,  5  +  5  +  9  +  1  +  1+7+9+2 
+  7  +  9  =  55,  the  mystic  number. 


A    CURIOUS    GEOMETRICAL    COINCIDENCE. 

If  an  equilateral  triangle  be  constructed  of  any 
size  and  a  circle  be  inscribed  therein  ;  and  towards 
each  vertex  of  the  triangle  additional  circles  be  in- 
scribed which  are  to  the  preceding  circle  as  3  to  7, 
with  -|-  of  their  diameters  and  -|  of  their  circumferences 
cut  by  the  larger  circle,  and  so  on  in  the  same  ratio 
to  infinity,  each  circle  will  touch  the  triangle,  and 
none  will  pass  its  bounds. 


v r~*> 

YD 


4 


